a. The particle has position vector
[tex]\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath[/tex]
[tex]\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath[/tex]
b. Its velocity vector is equal to the derivative of its position vector:
[tex]\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath[/tex]
c. At [tex]t=7.00\,\mathrm s[/tex], the particle has position
[tex]\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath[/tex]
[tex]\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m[/tex]
That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of [tex]\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m[/tex] away from the origin in a direction of [tex]\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ[/tex] relative to the positive [tex]x[/tex] axis.
d. The speed of the particle at [tex]t=7.00\,\mathrm s[/tex] is the magnitude of the velocity at this time:
[tex]\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath[/tex]
[tex]\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}[/tex]
Then its speed at this time is
[tex]\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}[/tex]
(a) The vector position of the particle at any time t; is s(t) = (8i)t + (j)t²
(b) the velocity of the particle at any time is v = 8i + (2j)t
(c) the position of the particle at t = 7.00 s is 56i + 49 j at a distance of 74.4 m
(d) the speed of the particle at t = 7.00 is 16.12 m/s
The given parameters;
acceleration of the particle, a = 2 j m/s²
initial velocity of the particle, v = 8 i m/s
(a) The vector position of the particle at any time t;
[tex]s = v_0t + \frac{1}{2} at^2\\\\s(t) = (8 \ i) t \ + (0.5 \times 2 \ j)t^2\\\\s(t) = (8 \ i)t \ + (j)t^2[/tex]
(b) the velocity of the particle at any time
[tex]velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\s' = v = 8i \ + (2j)t[/tex]
(c) the position of the particle at t = 7.00 s
[tex]s(t) = (8 \ i)t \ + (j)t^2\\\\s(7) = (8 \ i)\times 7 \ \ + \ \ (j)\times 7^2\\\\s(7) = 56i \ \ + \ \ 49j\\\\|s| = \sqrt{(56)^2 + (49)^2} = 74.4 \ m\\\\|s| = 74.4 \ m[/tex]
(d) the speed of the particle at t = 7.00
[tex]v(t) = 8i + (2j)t\\\\v(7) = 8i \ + (2j \times 7)\\\\v(7) = 8i + 14j\\\\|v| = \sqrt{8^2 + 14^2} \\\\|v| = \sqrt{260} \\\\|v| = 16.12 \ m/s[/tex]
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Four20 Ohm resistors are connected in parallel and the combination is connected to a 20-v emf device. The current in any one of the resistors is: A) 0.25 A B) 4.0 A C) 5.0 A D) 100 A E) L0A
Answer:
The current in any one of the resistors is 1 A.
Explanation:
It is given that, four 20 ohm resistors are connected in parallel. In parallel combination of resistors, the voltage across each and every resistor is same while the current divides. The equivalent resistance of all resistors is given by :
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}[/tex]
[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}+\dfrac{1}{20\ \Omega}[/tex]
[tex]R_{eq}=5\ \Omega[/tex]
Current flowing in the entire circuit can be calculated using Ohm's law as :
Current, [tex]I=\dfrac{V}{R_{eq}}[/tex]
[tex]I=\dfrac{20\ V}{5\ \Omega}[/tex]
I = 4 A
Since, in parallel combination current divides. So, current flowing in all four resistor divides and is 1 A. Hence, this is the required solution.
The current in a single 20 Ohm resistor connected in parallel to a 20 V emf device is 1 A. This is calculated using Ohm's law (I = V/R).
Explanation:The question involves calculating the current in a single resistor when four 20 Ohm resistors are combined in parallel and connected to a 20 V emf device. For resistors in parallel, they share the same potential difference, or voltage. In this case, the emf device provides each parallel resistor with a potential difference of 20 V.
By using Ohm's law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points, the current can be calculated. Ohm's law is typically represented by the formula I = V/R, where I is the current, V is the voltage, and R is the resistance. In this scenario, we can calculate the current as I = 20V/20Ohm which equals 1 A.
So, the current that will flow through any one of the 20 Ohm resistors connected in parallel and connected to a 20 V power supply would be 1 A. The correct answer falls outside the options given and is 1 A.
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The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in 1981. Assuming the initial launch angle was 45° and neglecting air resistance, answer the following. (a) Determine the initial speed of the projectile. m/s
Answer:
28.2 m/s
Explanation:
The range of a projectile launched from the ground is given by:
[tex]d=\frac{v^2}{g}sin 2\theta[/tex]
where
v is the initial speed
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\theta[/tex] is the angle at which the projectile is thrown
In this problem we have
d = 81.1 m is the range
[tex]\theta=45^{\circ}[/tex] is the angle
Solving for v, we find the speed of the projectile:
[tex]v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s[/tex]
A pendulum with a length of 1.5 meters is released from an angle of 20 degrees.What is the period and frequency of this pendulum?
Answer:
The period and frequency of this pendulum are 2.457 s and 0.407 Hz.
Explanation:
Given that,
Length = 1.5 m
Angle = 20°
We need to calculate the period
Using formula of period
[tex]T = 2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where, T = time period
g = acceleration due to gravity
l = length
Put the value into the formula
[tex]T=2\times3.14\sqrt{\dfrac{1.5}{9.8}}[/tex]
[tex]T=2.457\ sec[/tex]
We need to calculate the frequency
[tex]T = \dfrac{1}{f}[/tex]
[tex]f=\dfrac{1}{T}[/tex]
Put the value of T
[tex]f=\dfrac{1}{2.457}[/tex]
[tex]f =0.407\ Hz[/tex]
Hence, The period and frequency of this pendulum are 2.457 s and 0.407 Hz.
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball traveling horizontally at 37.0 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction.)
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)
(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s
(b) The work that the racket does on the ball is -35.1 Joule
Further ExplanationGiven :
mass of ball = m = 0.06 kg
initial velocity = v₁ = -50.4 m/s
final velocity = v₂ = 37.0 m/s
Unknown :
(a) Impulse = I = ?
(b) Work = W = ?
Solution :
Question (a) :In this question , we could use the formula from Second Law of Newton :
[tex]I = \Delta p[/tex]
[tex]I = p_2 - p_1[/tex]
[tex]I = m \times v_2 - m \times v_1[/tex]
[tex]I = m \times (v_2 - v_1)[/tex]
[tex]I = 0.06 \times (37.0 - (-50.4))[/tex]
[tex]I = 0.06 \times (87.4)[/tex]
[tex]I = 5.244~kg.m/s[/tex]
[tex]\large { \boxed {I \approx 5.24~kg.m/s} }[/tex]
Question (b) :[tex]W = F \times d[/tex]
[tex]W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)[/tex]
[tex]W = \frac{I(v_1 + v_2)}{2}[/tex]
[tex]W = \frac{5.244(-50.4 + 37)}{2}[/tex]
[tex]W = \frac{5.244(-13.4)}{2}[/tex]
[tex]W = -35.1348~Joule[/tex]
[tex]\large { \boxed {W \approx -35.1~Joule}}[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Dynamics
Keywords: Newton, Law, Impulse, Work
In this physics problem, we calculate the impulse delivered by the racket to the tennis ball and the work done by the racket on the ball.
The impulse delivered to the ball by the racket:
Calculate the change in momentum (final momentum - initial momentum).Impulse = Change in momentum.The work done by the racket on the ball:
Work = Change in kinetic energy.Consider the direction of the velocity to determine the sign of the work.If you drop an object from a height of 1.4 m, it will hit the ground in 0.53 s. If you throw a baseball horizontally with an initial speed of 35 m/s from the same height, how long will it take the ball to hit the ground?
Answer:
The ball to hit the ground in 0.53 s.
Explanation:
Given that,
Height = 1.4 m
Time t = 0.53 s
Initial speed = 35 m/s
We need to calculate the time when the ball to hit the ground
Using equation of motion
[tex]s_{y}=u_{y}t-\dfrac{1}{2}gt^2+h_{0}[/tex]
Where, s= vertical height
u= vertical velocity
t = time
h = height
Put the value in equation
[tex]0=0-\dfrac{1}{2}\times9.8\times t^2+1.4[/tex]
[tex]t^2=\dfrac{1.4}{4.9}[/tex]
[tex]t=\sqrt{\dfrac{1.4}{4.9}}[/tex]
[tex]t=0.53\ s[/tex]
Hence, The ball to hit the ground in 0.53 s.
Final answer:
A baseball thrown horizontally with an initial speed will take the same amount of time to fall as a dropped object from the same height, which is 0.53 seconds, because the horizontal speed does not affect the vertical fall time.
Explanation:
The time it takes for an object to fall from a height solely depends on the force of gravity and the initial vertical speed. Since the baseball is thrown horizontally with an initial speed of 35 m/s, this speed does not affect the vertical fall time. The ball will hit the ground in the same duration as any object dropped from the same height without any initial vertical velocity, provided that air resistance is negligible. The previously stated object took 0.53 s to fall from a height of 1.4 m, therefore the baseball will also take 0.53 seconds to hit the ground.
An electron traveling horizontally to the right enters a region where a uniform electric field is directed downward. What is the direction of the electric force exerted on the electron once it has entered the electric field?
Answer:
Upward
Explanation:
For charged particles immersed in an electric field:
- if the particle is positively charged, the direction of the force is the same as the direction of the electric field
- if the particle is negatively charged, the direction of the force is opposite to the direction of the electric field
In this problem, we have an electron - so a negatively charged particle - so the direction of the force is opposite to that of the electric field.
Since the electric field is directed downward, therefore, the electric force on the electron will be upward.
In which of the following pairs do both numbers contain the same number of significant figures? Check all that apply. Check all that apply. 0.00012 s and 12000 s 11.0 m and 11.00 m 0.0250 m and 0.205 m 250.0 L and 2.5×10−2L
Answer:
The correct answer is :A and C
Explanation:
According to the rule
1- Zeroes at the beginning of a number are never significant.
2- Zeroes at the end of a number are significant if there is a decimal point.
(a) 0.00012 s and 12000 s
0.00012 has 2 significant figure and 12000 has 2 significant figures
(b) 11.0 m and 11.00 m
11.0 has 3 significant figures and 11.00 has 4 significant figure.
(c) 0.0250 m and 0.205
0.0250 has 3 significant figure and 0.250 has 3 significant figures
(d) 250.0 L and 2.5×10−2L
250.0 has 4 significant figures and 2.5x10−2 has 2 sig figures.
So (a) and (c) pairs have same number of sig figures.
Final answer:
The pair out of the provided options that contains the same number of significant figures is 0.0250 m and 0.205 m, both with three significant figures.
Explanation:
When comparing significant figures in different numbers, we must apply the rules for determining the number of significant figures to each number.
0.00012 s has two significant figures (the leading zeros are not significant).
12000 s likely has two significant figures (unless further decimal places are indicated or if it's written in scientific notation, such as 1.2×10⁴ which would have three significant figures).
11.0 m and 11.00 m have three and four significant figures respectively (trailing zeros after a decimal are significant).
0.0250 m has three significant figures (leading zeros are not significant, but trailing zeros after a decimal are).
0.205 m has three significant figures (leading zeros are not significant).
250.0 L has four significant figures (trailing zeros after a decimal are significant).
2.5×10⁻² L has two significant figures (in scientific notation, all digits are significant).
Therefore, the pairs that contain the same number of significant figures are 0.0250 m and 0.205 m, both with three significant figures.
A centrifuge in a medical laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates through 54.0 revolutions before coming to rest. Find the constant angular acceleration (in rad/s2) of the centrifuge. Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. rad/s2
Answer:
-220 rad/s²
Explanation:
ω² = ω₀² + 2α(θ - θ₀)
We're given that ω = 0 rev/min, ω₀ = 3700 rev/min, and θ - θ₀ = 54.0 rev.
Converting to rad/s:
0 rev/min × (2π rad / rev) × (1 min / 60 s) = 0 rad/s
3700 rev/min × (2π rad / rev) × (1 min / 60 s) = 123⅓π rad/s
54.0 rev × (2π rad / rev) = 108π rad
Plugging in:
(0 rad/s)² = (123⅓π rad/s)² + 2α(108π)
α = -221.2 rad/s²
There's 2 significant figures in 3700 rev/min, so we need to round our answer to 2 sig figs. The angular acceleration is -220 rad/s².
A sample of gas in a balloon has an initial temperature of 23 ∘C and a volume of 1.09×103 L . If the temperature changes to 59 ∘C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?
Answer:
[tex]1.22\cdot 10^3 L[/tex]
Explanation:
We can solve the problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas is directly proportional to its temperature:
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where here we have
[tex]V_1 = 1.09\cdot 10^3 L[/tex] is the initial volume
[tex]T_1 = 23^{\circ}+ 273 = 296 K[/tex] is the initial temperature
[tex]V_2[/tex] is the final volume
[tex]T_2 = 59^{\circ}+ 273 =332 K[/tex] is the final temperature
Solving for V2, we find
[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(1.09 \cdot 10^3 L)(332 K)}{296 K}=1.22\cdot 10^3 L[/tex]
What is the charge passing through a battery if 7.2A of current flows for 4.7 hours? 122 x 105 C 0 1.29 x 105 C 0 159 x 100c O 201 x 10s c
Answer:
The charge passing through a battery is [tex]1.22\times10^{5}\ C[/tex].
Explanation:
Given that,
Current = 7.2 A
Time = 4.7 hours
We need to calculate the charge
The charge is the product of current and time.
Using formula of charge
[tex]Q=it[/tex]
Where, Q = charge
i = current
t = time
Put the value into the formula
[tex]Q=7.2\times4.7\times3600[/tex]
[tex]Q=1.22\times10^{5}\ C[/tex]
Hence, The charge passing through a battery is [tex]1.22\times10^{5}\ C[/tex].
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 252 N in the negative x direction to try and move the 56 kg refrigerator. The coefficient of static friction is 0.61. (a) How much static frictional force does the floor exert on the refrigerator? Give both magnitude (in N) and direction. magnitude N direction (b) What maximum force (in N) do you need to apply before the refrigerator starts to move?
(a) 252 N, opposite to the applied force
There are two forces acting on the refrigerator in the horizontal direction:
- the pushing force of 252 N, F, forward
- the frictional force, Ff, pulling backward
In this case, the refrigerator is not moving: this means that its acceleration is zero. According to Newton's second law, this also means that the net force acting on the refrigerator is also zero:
[tex]\sum F = ma = 0[/tex]
So we have
[tex]F-F_f = 0[/tex]
which means that the frictional force is equal in magnitude to the pushing force:
[tex]F_f = F = 252 N[/tex]
and the direction is opposite to the pushing force.
(b) 334.8 N
The force that must be applied to the refrigerator to make it moving is equal to the maximum force of friction, which is given by:
[tex]F_{max} = \mu mg[/tex]
where
[tex]\mu = 0.61[/tex] is the coefficient of static friction
m = 56 kg is the mass of the refrigerator
g = 9.8 m/s^2 is the acceleration of gravity
Substituting:
F_max = (0.61)(56 kg)(9.8 m/s^2)=334.8 N
The force that will move the object is 83 N in the negative x direction.
What is frictional force?The frictional force is the force that opposes motion. The force of friction acts in the opposite direction to the force that is moving the object. Now we can obtain the frictional force from;
μs = F/mg
F = μsmg = 0.61 * 56 kg * 9.8 m/s^2 = 335 N (positive direction)
Secondly;
ma = F - Ff
Where;
ma = resultant force
F = moving force
Ff = frictional force
Now we need to find the net force ma
ma = 252 N - 335 N
ma = -83 N
The force that will move the object is 83 N in the negative x direction.
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A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102 MV. (a) What energy was dissipated? (b) What mass of water could be raised from 15ºC to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.
Answer:
part a)
[tex]U = 2\times 10^9 J[/tex]
Part b)
[tex]m = 5.6 \times 10^3 kg[/tex]
Part c)
Huge damage will be there in tree
Explanation:
Part a)
Energy loss due to charge flow through given potential difference is given as
[tex]U = qV[/tex]
now plug in all the values in it
[tex]U = (20 C)(1.00 \times 10^2 \times 10^6 V)[/tex]
[tex]U = 2\times 10^9 J[/tex]
Part b)
Now this energy is used to raise the temperature of water from 15 degree C to boiling point
So here we will have
[tex]Q = ms\Delta T[/tex]
[tex]2 \times 10^9 = m(4186)(100 - 15)[/tex]
[tex]m = 5.6 \times 10^3 kg [/tex]
Part c)
Since in part b) we can say its a huge amount of water that will boil due to the amount of energy that strikes to the tree.
So it will make huge damage to the tree
Final answer:
A lightning bolt moving 20.0 C of charge through a potential difference of 1.00 × 10² MV dissipates 2.00 × 10¹ joules of energy. This energy can boil a significant mass of water, demonstrating the power of lightning. The rapid boiling of sap and expansion of steam can cause the tree to explosively splinter.
Explanation:
Understanding the Physics of a Lightning Strike on a Tree
A lightning bolt striking a tree involves complex physical phenomena, including the movement of charge and the dissipation of significant amounts of energy. When a lightning bolt with 20.0 C of charge passes through a potential difference of 1.00 × 10² MV, we can calculate the energy dissipated using the relationship between charge, potential difference, and energy.
(a) Calculating Dissipated Energy:
The energy dissipated can be calculated using the formula E = QV, where E is the energy in joules, Q is the charge in coulombs, and V is the potential difference in volts. Given the charge (20.0 C) and the potential difference (1.00 × 10² MV or 1.00 × 10¸ V), the energy dissipated is 2.00 × 10¹ joules (or 2 billion joules).
(b) Heating and Boiling Water with This Energy:
To determine what mass of water could be raised from 15°C to the boiling point and then boiled, we need to consider the specific heat capacity of water and the latent heat of vaporization. However, without diving into specific calculations here, it's clear that the energy amount is sufficient to boil a significant mass of water, showcasing the tremendous power of lightning.
(c) Potential Damage to the Tree:
The sudden energy release and the boiling of sap inside the tree can cause severe damage. The rapid expansion of steam may result in the explosive splintering of the tree trunk, a common effect seen in trees struck by lightning.
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of light to get from the Sun to the Earth (in s) is written in scientific notation. Note: the speed of light is approximately 3.0×108 m/s.
The time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]
The formula for calculating the average speed is expressed according to the formula:
[tex]Speed= \frac{distance}{time}[/tex]
Given the speed of light as [tex]3.0 \times 10^8m/s[/tex]
Distance from earth to the sum is [tex]149.6\times 10^9m[/tex]
Substitute the given parameters into the formula to get the time "t"
[tex]t=\frac{d}{t} \\t=\frac{149.6\times10^9m}{3.0\times10^8}\\t=49.87\times 10\\t =498.7secs[/tex]
Hence the time it takes a beam of light to get from the Sun to the Earth is [tex]4.99\times 10^2secs[/tex]
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A small but dense 2.0-kg stone is attached to one end of a very light rod that is 1.2 m long. The other end of the rod is attached to a frictionless pivot. The rod is raised until it is vertical, with the stone above the pivot. The rod is released and the stone moves in a vertical circle with no air resistance. What is the tension in the rod as the stone moves through the bottom of the circle? (a) 20 N (b) 40 N (c) 60 N (d) 80 N (e) 100 N
The tension in the rod as the stone moves through the bottom of the circle is 80 N, which is found by applying the principles of conservation of energy and circular motion.
Explanation:The problem involves the concept of centripetal force and the principle of conservation of energy. When the stone is at the bottom of the circular path, the centripetal force necessary to keep it moving in a circle is provided by the tension, T in the rod and the weight of the stone. The tension can be calculated using the formula for the centripetal force, which is F = m * g + m * v²/r where m is the mass of the stone (2 kg), g is the acceleration due to gravity (approximate 10 m/s²), v is the velocity of the stone and r is the radius of of the circle.
Since the energy at the top of the circle (potential energy) is equal to the energy at the bottom of the circle (kinetic energy + potential energy), and we know that potential energy = m * g * h and kinetic energy = 1/2 * m * v², we can solve for v² = 2 * g * h. Hence, substituting v² in the previous formula, we get T = m * g + m * (2 * g * h)/r. Considering that h=r=1.2m, this simplifies to T = 4 * m * g , which, when evaluated equals 80N. Therefore, the correct answer is (d) 80 N.
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If you fell 90ft while sitting down strapped in an elevator (think tower of terror ride) would you die when the elevator hit the floor? Would the power be enough? Not for school just curious
Probably not. The elevator has security system just like any device. It will automatically trigger breaks in case it detects some sort of free fall.
Person inside free falling elevator could experience 0G or no gravity effect something similar to what astronauts experience in earth's orbit.
The only damage that could happen to you is when the breaks are released you won't feel 0G anymore and fall about 4ft to the floor of the elevator.
Hope this helps.
r3t40
An object is dropped from a tower, 400 ft above the ground. The object's height above ground t seconds after the fall is s(t)equals400 minus 16 t squared. Determine the velocity and acceleration of the object the moment it reaches the ground. The velocity of the object the moment it reaches the ground is nothing ft/s.
Answer: v= 160ft/s
a=32ft/s^2 constant
Explanation:
s(t)=400-16t^2 derivative of position is velocity v(t) and derivative of velocity is acceleration a(t) so let s(t)=0 to find the time of flight to reach the ground and take the two derivatives and use the time found and solve. Also acceleration is a constant as it’s gravity.
0=400-16t^2
400=16t^2
25=t^2
t=5s
ds/dt=v(t)=0-32t
dv/dt=a(t)=-32 constant(gravity)
v(t)=-32(5s)= -160ft/s negative sign is only showing direction
The velocity of the object the moment it reaches the ground is 160 ft/s.
What is Velocity?Velocity is defined as the directional speed of an object.How to solve this problem?The problem can be solved by following steps.
s(t) = 400-16t^2 (given)
We know that derivative of position is Velocity v(t) and the derivate of velocity is acceleration a(t)Let s(t)=0
Therefore
0= 400-16t^2
400=16t^2
25=t^2
Therefore
t = 5sec
Now as we know that the derivative of the position is Velocity
so v(t) = ds/dt = -32t
where t = 5sec
substitute the value of t in v(t)
Therefore, v(t) = -32(5) = -160
The direction is negative
Hence the velocity is 160ft/s
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Choose the letter for the acceleration of the ball during the upward motion after it is released. a) The acceleration is in the negative direction and constant. b) The acceleration is in the negative direction and increasing. c) The acceleration is in the negative direction and decreasing. d) The acceleration is zero. e) The acceleration is in the positive direction and decreasing. ea) The acceleration is in the positive direction and increasing. ea) The acceleration is in the positive direction and constant
Answer:
Option (a)
Explanation:
If a body is thrown upwards, it's velocity goes on decreasing with constant rate. It is because an acceleration is acting on the body which is equal to acceleration due to gravity and acting downwards. The value of acceleration due to gravity is constant and always acting downwards.
A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. Remember, the density of water is 1.0 g/m
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Data obtained from the question
Mass of copper (M꜀) = 248 g
Volume of water = 390 mL
Density of water = 1 g/mL
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =? Step 2:Determination of the mass of water
Volume of water = 390 mL
Density of water = 1 g/mL
Mass of water =?[tex]Density = \frac{mass}{volume}\\\\1 = \frac{mass}{390}[/tex]
Cross multiply
[tex]Mass = 1 * 390[/tex]
Mass of water = 390 gStep 3:Determination the initial temperature of the copper.
Mass of copper (M꜀) = 248 g
Mass of water (Mᵥᵥ) = 390 g
Initial temperature of water (Tᵥᵥ) = 22.6 °C
Equilibrium temperature (Tₑ) = 39.9 °C
Initial temperature of copper (T꜀) =? NOTE:1. Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC
2. Specific heat capacity of copper (C꜀) = 0.385 J/gºC
Heat lost by copper = heat gained by water
[tex]Q_{c} = Q_{w} \\ \ M_{c} C_{c}(T_{c}-T_{e}) = M_{w} C_{w}(T_{e}-T_{w})\\248* 0.385(T_{c}-39.9) = 390*4.184(39.9-22.6)\\95.48(T_{c}-39.9) = 1631.76*17.3\\95.48(T_{c}-39.9) = 28229.448[/tex]
Divide both side by 95.48
[tex]T_{c} - 39.9 = \frac{28229.448}{95.48}\\T_{c} - 39.9 = 295.658[/tex]
Collect like terms
[tex]T_{c} = 295.658 + 39.9[/tex]
T꜀ = 335.6 °CTherefore, the initial temperature of the piece of copper is 335.6 °C.
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A satellite moves in a circular orbit a distance of 1.6×10^5 m above Earth's surface. (radius of Earth is 6.38 x 10^6m and its mass is 5.98 x 10^24 kg). Determine the speed of the satellite.
Answer:
The speed of the satellite is 7809.52 m/s
Explanation:
It is given that,
Radius of Earth, [tex]r=6.38\times 10^6\ m[/tex]
Mass of earth, [tex]M=5.98\times 10^{24}\ kg[/tex]
A satellite moves in a circular orbit a distance of, [tex]d=1.6\times 10^5\ m[/tex] above Earth's surface.
We need to find the speed of the satellite. It is given by :
[tex]v=\sqrt{\dfrac{GM}{R}}[/tex]
R = r + d
[tex]R=(6.38\times 10^6\ m+1.6\times 10^5\ m)=6540000\ m[/tex]
So, [tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 5.98\times 10^{24}\ kg}{6540000\ m}}[/tex]
v = 7809.52 m/s
So, the speed of the satellite is 7809.52 m/s. Hence, this is the required solution.
A 1000-kg car is traveling north at 20.0 m/s. A 1500-kg car is traveling north at 36.0 m/s. The 1500-kg collides with the rear of the 1000- kg car and they lock together. Ignoring external forces acting during the collision, what is the velocity of the cars immediately after the collision? (a) 56.0 m/s north (b) 29.6 m/s north (c) 13.6 m/s south (d) 8.00 m/s south (e) 28.0 m/s north
Answer:
B
Explanation:
Use conservation of momentum: Total momentum of the system before is the same as the total momentum after the collision. Since both are moving north, the momentum of both cars is in the same direction, and the total momentum before collision is 1000kg*20m/s + 1500kg*36m/s = 74000kg-m/s.
This is the same amount of momentum after collision, and since they are locked together, their mass is added. The velocity can be found by: 74000kg-m/s ÷ 2500kg = 29.6m/s.
momentum = mass*velocity
The velocity of the two cars combined, immediately after the collision, can be determined by conserving the total momentum before the collision. This velocity is 29.6 m/s north.
Explanation:This problem is about the conservation of momentum in a collision. Momentum before the collision is the total of the momentum of the two cars combined. We can calculate this using the formula 'momentum = mass x velocity'. For the 1000-kg car it's (1000 kg)x(20 m/s) = 20000 kg.m/s and for the 1500-kg car, it's (1500 kg)x(36 m/s) = 54000 kg.m/s. The total momentum before the collision is 20000 + 54000 = 74000 kg.m/s.
After the collision, the two cars move together, meaning the total mass is now 1000 + 1500 = 2500 kg. The question asks for the velocity (v) of the cars immediately after the collision. As the total momentum must be conserved, we can find this velocity using the formula 'momentum = total mass x velocity'. So, 74000 = (2500 kg) x v. To solve for v, we divide 74000 by 2500. Doing this gives v = 29.6 m/s. Therefore, the velocity of the cars immediately after the collision is 29.6 m/s north.
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what net force is needed to give a 48.0kg grocery cart an acceleration of 3.39m/s^2 ?
you use the formula F=ma to get the answer
Which style of parenting is most likely to produce spoiled immature and self indulgent children who have little self control?
The style of parenting is most likely to produce self indulgent children who have little self control.
A biological parent is the person who contributed to the child's genes, and in the case of the mother, carried the child during a pregnancy. A biological parent can, but doesn't have to be a legal guardian of the child.
An adoptive parent is did not directly contribute to the child's genes, but took over the care of the child after the child's birth. An adoptive parent is always a legal guardian of the child and maintains the contact with the child after their maturity.
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Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person wearing a g-suit can withstand accelerations up to about 9g (88.2 m/s2) without losing consciousness. HINT (a) If a human centrifuge has a radius of 5.15 m, what angular speed (in rad/s) results in a centripetal acceleration of 9g? rad/s (b) What linear speed (in m/s) would a person in the centrifuge have at this acceleration? m/s
(a) [tex]4.14 rad/s^2[/tex]
The relationship beween centripetal acceleration and angular speed is
[tex]a=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
r is the radius of the circular path
Here we gave
[tex]a = 9g = 88.2 m/s^2[/tex] is the centripetal acceleration
r = 5.15 m is the radius
Solving for [tex]\omega[/tex], we find:
[tex]\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2[/tex]
(b) 21.3 m/s
The relationship between the linear speed and the angular speed is
[tex]v=\omega r[/tex]
where
v is the linear speed
[tex]\omega[/tex] is the angular speed
r is the radius of the circular path
In this problem we have
[tex]\omega=4.14 rad/s[/tex]
r = 5.15 m
Solving the equation for v, we find
[tex]v=(4.14 rad/s)(5.15 m)=21.3 m/s[/tex]
Explanation:
(a) Centripetal acceleration, [tex]a=9g=88.2\ m/s^2[/tex]
Radius, r = 5.15 m
Let [tex]\omega[/tex] is the angular speed. The relation between the angular speed and angular acceleration is given by :
[tex]a=\omega^2 r[/tex]
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
[tex]\omega=\sqrt{\dfrac{88.2}{5.15}}[/tex]
[tex]\omega=4.13\ rad/s[/tex]
(b) Let v is the linear speed of the person in the centrifuge have at this acceleration. It is given by :
[tex]v=r\times \omega[/tex]
[tex]v=5.15\times 4.13[/tex]
v = 21.26 m/s
Hence, this is the required solution.
The largest country in the world is Russia with a land area of 16,376,870 km2 Wh area of the Russia in m' 11. A) 1.637687 x10 B) 1.637687 x10 C) 1.637687 x10 D) 1637687 x10" E) 1.637687 x10
Answer:
1.637687 x 10¹³ m²
Explanation:
Land area of Russia has been given in the question as
A = land area of Russia = 16,376,870 km²
we know that
1 km = 1000 m
hence
A = land area of Russia = 16,376,870 (1000 m)² = 16,376,870,000,000 m²
A = 16,376,870,000,000 m²
To represent the area in exponent form, we shift the decimal place from right to left until we have one decimal place.
So we get
A = 1.637687 x 10¹³ m²
How many electrons does it take to make up 4.33 C of charge?
Answer:
Number of electrons, [tex]n=2.7\times 10^{19}[/tex]
Explanation:
It is given that,
Charge, q = 4.33 C
We need to find the number of electrons that make 4.33 C of charge. According to quantization of charge as :
[tex]q=ne[/tex]
n = number of electrons
e = electron's charge
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{4.33\ C}{1.6\times 10^{-19}\ C}[/tex]
[tex]n=2.7\times 10^{19}[/tex]
So, the number of electrons are [tex]2.7\times 10^{19}[/tex] Hence, this is the required solution.
An electrical device draws 4.68 A at 220 V. (a) If the voltage drops by 31%, what will be the current, assuming the resistance doesn't change?
Answer:
The current will be 3.23 A.
Explanation:
Given that,
Current I = 4.68 A
Voltage V = 220 volt
We need to calculate the resistance
Using ohm's law
[tex]V = I R[/tex]
[tex]R = \dfrac{V}{I}[/tex]
Where,
V = voltage
I = current
R = resistance
Put the value into the formula
[tex]R = \dfrac{220}{4.68}[/tex]
[tex]R = 47\ \Omega[/tex]
We need to calculate the current
If the voltage drops by 31%
Voltage will be
[tex]V'=V-V\times31%[/tex]
[tex]V'=220-220\times\dfrac{31}{100}[/tex]
[tex]V'=151.8\ volt[/tex]
Now, the current will be
[tex]I = \dfrac{151.8}{47}[/tex]
[tex]I=3.23\ A[/tex]
Hence, The current will be 3.23 A.
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance. What is its vertical displacement (in m) after 6 s?
Answer:
176.4 m
Explanation:
U = 0, t = 6s, g = 9.8 m/s^2
Use second equation of motion
H = ut + 1/2 gt^2
H = 0 + 0.5 × 9.8 × 6 × 6
H = 176.4 m
It is the displacement from the point of dropping of object.
Water flows through a multisectional pipe placed horizontally on the ground. The velocity is 3.0 m/s at the entrance and 2.1 m/s at the exit. What is the pressure difference between these two points? Neglect friction.
Answer:
[tex]P_1 - P_2 = 2295 pascal[/tex]
Explanation:
As we know by Bernoulli's principle
[tex]P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2[/tex]
[tex]P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2} \rho v_1^2[/tex]
here we know that
[tex]\rho = 1000 kg/m^3[/tex]
also we know that
[tex]v_1 = 2.1 m/s[/tex]
[tex]v_2 = 3 m/s[/tex]
now we have
[tex]P_1 - P_2 = \frac{1}{2}(1000)(3.0^2 - 2.1^2)[/tex]
[tex]P_1 - P_2 = 2295 pascal[/tex]
To find the pressure difference between two points in a pipe, we can use Bernoulli's equation. Bernoulli's equation states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
Explanation:To find the pressure difference between two points in a pipe, we can use Bernoulli's equation. Bernoulli's equation states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline. In this case, since the pipe is horizontal and we neglect friction, the potential energy term can be ignored. So we are left with:
P1 + 1/2ρv12 = P2 + 1/2ρv22
Where P1 and P2 are the pressures at the entrance and exit respectively, ρ is the density of the fluid, and v1 and v2 are the velocities at the entrance and exit respectively.
Using the given values, we can calculate the pressure difference.
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When an object vibrates in simple harmonic motion in the vertical direction, its minimum speed occurs when it _______.
is at the lowest pointis
at the equilibrium point
is at either endpoint
is at the highest point
The minimum speed of an object in simple harmonic motion occurs when it is at the equilibrium point. Hence the correct answer is option B
Explanation:In simple harmonic motion, the minimum speed of an object occurs when it is at the equilibrium point. The equilibrium position is where the object would naturally rest in the absence of force. When the object is at the equilibrium point, it has zero acceleration and the speed is at its minimum.
Hence the correct answer is option B
Would it be correct to say the Doppler Effect is the apparent change in the speed of a wave due to the motion of the source? Be sure to fully explain your position using at least 3 complete content related sentences.
Answer:
No, it is not correct to say the Doppler Effect is the apparent change in the speed of a wave due to the motion of the source.
Explanation:
While Doppler Effect is due to motion of the source or observer but it is not the apparent change in the speed of the wave. The speed of the wave remains the same. It is the wavelength and frequency of the waves that change in Doppler Effect when there is a relative motion between source and observer.
The Doppler Effect is the change in frequency, not speed, of a wave due to the relative motion between the source and observer.
Explanation:It would not be correct to say the Doppler Effect is the apparent change in the speed of a wave due to the motion of the source. The Doppler Effect actually refers to the apparent change in the frequency of a wave as a result of the relative motion between the source of the wave and the observer. While the speed of the wave itself remains constant, the wavelength experienced by the observer can either compress or stretch depending on whether the source is moving towards or away from them, thereby changing its perceived frequency.